It's unlikely that an implementation using continuations would be
faster than one that does not.
An idiomatic solution might look like:
(define (map-once fn xs)
(for/list ([i (in-range (length xs))])
(for/list ([(x j) (in-indexed (in-list xs))])
(cond [(= i j) (fn x)]
[else x]))))
But it's not terribly fast.
If you're willing to use vectors instead of lists, then maybe:
(define (map-once fn xs)
(build-vector (vector-length xs)
(λ (i)
(define v (vector-copy xs))
(vector-set! v i (fn (vector-ref v i)))
v)))
On Fri, Jun 19, 2015 at 3:24 PM, Luke Miles <rashreportl...@gmail.com> wrote:
> Say I have a list ls and I want to produce a list of
> lists where the i'th list has the i'th element of ls tripled,
> but all other elements are the same.
>
> e.g. '(3 5 7) => '((9 5 7) (3 15 7) (3 5 21))
>
> What is a fast way to do this?
>
>
> I could do a loop with appending.
> (define (map-once f ls)
> (let M ([sooner null] [later ls])
> (if (null? later) null
> (cons (append sooner (list (f (car later))) (cdr later))
> (M (append sooner (list (car later))) (cdr later))))))
>
> -> (map-once sqr '(4 5 6))
> '((16 5 6) (4 25 6) (4 5 36))
>
> Unfortunately, this is very slow & messy.
> I have to do 2 big appends for every element is the return list.
>
>
> Here is a cleaner-looking, but still slow way:
> (define (list-set ls i new-val)
> (let-values ([(sooner later) (split-at ls i)])
> (append sooner (list new-val) (cdr later))))
>
> (define (map-once f ls)
> (for/list ([i (in-naturals)]
> [elm (in-list ls)])
> (list-set ls i (f elm))))
>
>
> I'm thinking a good implementation might use continuations somehow?
>
> Maybe of vector-set (without the exclamation point) existed, I could use it?
>
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